Step 1: List every minterm that reads $1$.
Reading the K-map row by row (rows $AB = 00,01,11,10$; columns $CD = 00,01,11,10$), the $1$-cells are at $(A,B,C,D)$ equal to $(0,1,0,1)$, $(0,1,1,1)$, $(0,1,1,0)$, $(1,1,0,0)$, $(1,1,1,1)$, $(1,1,1,0)$, $(1,0,1,0)$, i.e. minterms $m_5, m_6, m_7, m_{10}, m_{12}, m_{14}, m_{15}$.
Step 2: Find every prime implicant by pairing adjacent 1's.
$m_6$ and $m_7$ differ only in $D$, both have $A{=}0,B{=}1,C{=}1$: pair gives $\bar{A}BC$. $m_{14}$ and $m_{15}$ differ only in $D$, both have $A{=}1,B{=}1,C{=}1$: pair gives $ABC$. $\bar{A}BC$ and $ABC$ differ only in $A$, so they merge into the bigger group $BC$, a prime implicant covering $m_6, m_7, m_{14}, m_{15}$. $m_5$ and $m_7$ differ only in $C$, both have $A{=}0,B{=}1,D{=}1$: prime implicant $\bar{A}BD$, covering $m_5, m_7$. $m_{10}$ and $m_{14}$ differ only in $B$, both have $A{=}1,C{=}1,D{=}0$: prime implicant $AC\bar{D}$, covering $m_{10}, m_{14}$. $m_{12}$ and $m_{14}$ differ only in $C$, both have $A{=}1,B{=}1,D{=}0$: prime implicant $AB\bar{D}$, covering $m_{12}, m_{14}$.
Step 3: Build a coverage list.
$BC$ covers $m_6, m_7, m_{14}, m_{15}$. $\bar{A}BD$ covers $m_5, m_7$. $AC\bar{D}$ covers $m_{10}, m_{14}$. $AB\bar{D}$ covers $m_{12}, m_{14}$. $m_5$ is covered only by $\bar{A}BD$, $m_{10}$ only by $AC\bar{D}$, and $m_{12}$ only by $AB\bar{D}$, so these three are essential prime implicants. $BC$ is also essential since nothing else covers $m_6$ or $m_{15}$.
Step 4: Read off the minimum expression.
Taking the four essential prime implicants together covers every marked cell at minimum cost:
\[ F = BC + \bar{A}BD + AC\bar{D} + AB\bar{D} \]
Step 5: Checking the alternatives.
Option (B) replaces some of these clean prime implicants with longer 4-literal terms such as $AB\bar{C}\bar{D}$, which cost more literals than needed for the same cells, so it fails the minimality test. Option (C) swaps in $AB\bar{C}\bar{D}$, a term covering only $m_{12}$, in place of the cheaper 2-cell group $AB\bar{D}$ that also picks up $m_{14}$, so it is not minimal either. Option (D) throws in $BCD$, a term fully contained inside the already-used $BC$, which adds nothing and breaks minimality.
Final Answer:
\[ F = BC + \bar{A}BD + AC\bar{D} + AB\bar{D} \]
This matches option (A).